hard logarithm question help
the tangent to the graph y= e to the kx at x=3 has equation of e^b x y + 4x=13 find the values of k and b
I tried to use dy/dx of y=e to the kx and got dy/dx= ke to the kx and tried to sub in 3 for the question, then i tried to for the other equation to substitute y as mx+c and substitute the gradient of the other graph in this equation but am still confused
I tried to use dy/dx of y=e to the kx and got dy/dx= ke to the kx and tried to sub in 3 for the question, then i tried to for the other equation to substitute y as mx+c and substitute the gradient of the other graph in this equation but am still confused
Original post by dwrfwrw
the tangent to the graph y= e to the kx at x=3 has equation of e^b x y + 4x=13 find the values of k and b
I tried to use dy/dx of y=e to the kx and got dy/dx= ke to the kx and tried to sub in 3 for the question, then i tried to for the other equation to substitute y as mx+c and substitute the gradient of the other graph in this equation but am still confused
I tried to use dy/dx of y=e to the kx and got dy/dx= ke to the kx and tried to sub in 3 for the question, then i tried to for the other equation to substitute y as mx+c and substitute the gradient of the other graph in this equation but am still confused
Can you upload your working?
Original post by dwrfwrw
my data on my phone is quite slow so durijng my working all I have done ke to the kx at x =3 so i sub'ed in 3 so got dy/dx= ke to the 3k then got quite confused on how to apply this to solvefor k and b
You must realise that your post/formatting are hard to read?
What is the given equation of the tangent?
Why not upload the question/working when it's more convenient?
(edited 3 years ago)
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